Problem: Let $R$ be the region enclosed by the line $y=\dfrac x2$, the curve $y=2\sqrt x$, and the line $x=9$. $y$ $x$ ${y=2\sqrt x}$ ${y=\dfrac x2}$ $y=6}$ $ 9$ $ 0$ $ R$ A solid is generated by rotating $R$ about the line $y=6$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\pi \int_0^9 \left( 4x-\dfrac{x^2}{4} \right)dx$ (Choice B) B $\pi \int_0^9 \left[36-\left(2\sqrt x-\dfrac x2 \right)^2 \right]dx$ (Choice C) C $\pi \int_0^9 \left( \dfrac{x^2}{4}-10x+24\sqrt x \right)dx$ (Choice D) D $\pi \int_0^9 \left( \dfrac{x^2}{4}+4x-36 \right)dx$
Solution: Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=2\sqrt x}$ ${y=\dfrac x2}$ $y=6}$ $ 9$ $ 0$ Let the thickness of each slice be $dx$, let the radius of the washer, as a function of $x$, be $r_1(x)$, and let the radius of the hole, as a function of $x$, be $r_2(x)$. Then, the volume of each slice is $\pi[(r_1(x))^2-(r_2(x))^2]\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [(r_1(x))^2-(r_2(x))^2]\,dx$ This is called the washer method. What we now need is to figure out the expressions of $r_1(x)$ and $r_2(x)$, and the interval of integration. $r_1(x)$ is equal to the distance between the line $y=\dfrac x2$ and the line $y=6$. So, ${r_1(x)=6-\dfrac x2}$. $r_2(x)$ is equal to the distance between the curve $y=2\sqrt x$ and the line $y=6$. So, ${r_2(x)=6-2\sqrt x}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [({r_1(x)})^2-({r_2(x)})^2] \\\\ &= \pi\left[ \left( {6-\dfrac x2} \right)^2-\left( {6-2\sqrt x} \right)^2 \right] \\\\ &=\pi\left[ \left( 36-6x+\dfrac{x^2}{4} \right)-\left(36-24\sqrt x+4x \right) \right] \\\\ &=\pi\left( \dfrac{x^2}{4}-10x+24\sqrt x \right) \end{aligned}$ The leftmost endpoint of $R$ is at $x=0$ and the rightmost endpoint is at $x=9$. So the interval of integration is $[0,9]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^9 \pi\left( \dfrac{x^2}{4}-10x+24\sqrt x \right)dx \\\\ &=\pi \int_0^9 \left( \dfrac{x^2}{4}-10x+24\sqrt x \right)dx \end{aligned}$